solve for y in first equation x+y=16 y=16-x subsitiute that for y in other equation
x^2+(16-x)^2=sum x^2+x^2-32x+256=sum 2x^2-32x+256=sum take derivitive to find the minimum value (or just find the vertex because the parabola opens up) derivitive is 4x-32=derivitve of sum the max/min is where the derivitive equals 0 4x-32=0 4x=32 x=8
